So I am new to Haskell and below I have attempted to program a function that takes a given value of e and a given list and determines whether that given value appears in the list given outputting True if the value given does appear and False if not.
inListm e [] = False
inListm e (x:xs)
| e == x = True || inListm e xs
| otherwise = False || inListm e xs
If
inListm 2 [0, 2, 1, 2]
is given, the output would be
True
However, I would like the final output to be in a list like this
[True]
I have attempted to do this by
inListd e [] = False : []
inListd e (x:xs)
| e == x = True : [] || inListd e xs
| otherwise = False :[] || inListd e xs
but all that gives me is an error so I would like to know how I could resolve this
Consider reusing your existing function; reuse is a big part of what makes writing big programs possible. Like this:
inListd e xs = inListm e xs : []
-- OR
inListd e xs = [inListm e xs]
You are on the right track. The only thing necessary is to return a list if you have a result, so:
inListd :: Eq a => a -> [a] -> [Bool]
inListd e [] = [False]
inListd e (x:xs)
| e == x = [True]
| otherwise = inListd e xs
That being said, I does not seem to make much sense to wrap the result in a list.
Related
I'm trying to test lists for duplicates but when I compile and enter
repeated [1,2,3,4]
it outputs True. What's wrong?
belongs :: Eq a => a -> [a] -> Bool
belongs n [] = False
belongs n (x:xs) | n == x = True
| otherwise = belongs n xs
repeated :: [Integer] -> Bool
repeated [] = False
repeated (x:xs) | belongs x xs = True
| otherwise = belongs (head xs) xs
"belongs (head xs) xs" checks whether xs's head is within xs, which will always be true.
(Except when xs is empty, in which case your program will crash! "head" is a partial function, crashes with empty lists)
This would fix it (as pointed by #talex too, but I also suggest making it more general, there is no need to specialize it to Integer):
repeated :: Eq a => [a] -> Bool
repeated [] = False
repeated (x:xs) | belongs x xs = True
| otherwise = repeated xs
You want
repeated :: [Integer] -> Bool
repeated [] = False
repeated (x:xs) | belongs x xs = True
| otherwise = repeated xs
Background
We are implementing this algorithm in F#.
Here is a little bit more information from Topor (1982) about the notation that the algorithm uses:
Formally, a 't list is either null (denoted nil) or has a hd (which is a 't) and a tl (which is a 't list)... If x is a list, we test whether it is null by writing null x... We create a new list, adding the element a at the front of an existing list x, by writing a:x... We denote the unit list containing the element a by list(a)... list(x) = x:nil.
Question
What we're wondering is how in F# to express those nil, null, and list(nil) values. For instance, should we be using the Option type, an empty list, or something else?
What We Have Tried
let rec kpermute k (xs: 't list) =
let rec mapPerm k xs ys =
match ys with
| [] -> []
| head::tail ->
let kpermuteNext = kpermute (k-1) (removeFirst head xs)
let mapPermNext = mapPerm k xs tail
mapcons head kpermuteNext mapPermNext
match k with
| 0 -> [[]]
| _ when xs.Length < k -> []
| _ -> mapPerm k xs xs
When working with lists, for list(nil) we use [[]] and for nil we use []. While that's fine, there might be a more expressive way to do it. There are also times when we use List.empty<'t list> and List.empty<'t> when the type inference needs more information.
The paper gives you all the answers: nil is []; null x is a test for whether x is the empty list; list(nil) is [[]].
The naïve translation of algorithm B to F# is as follows:
let rec minus a = function
| [] -> failwith "empty list"
| xh :: xt -> if xh = a then xt else xh :: minus a xt
let rec permute2 k x =
if k = 0 then [[]]
elif List.length x < k then []
else mapperm k x x
and mapperm k x = function
| [] -> []
| yh :: yt -> mapcons yh (permute2 (minus yh x)) (mapperm x yt)
and mapcons a ps qs =
match ps with
| [] -> qs
| ph :: pt -> a :: ph :: mapcons a pt qs
I am trying to fill ma lazylist by unpaired elements (with recursion), starting with element k. For example: k = 2, list is [2,3,5,7,9,...] The code:
let lgen =
let rec gen k = LCons(k, fun () -> gen k (k + 2))
in gen 1;;
But how can I check is the element k unpaired? (I think that here I need to use match).
Assuming your type for lazy lists is something like this:
type 'a llist = LNil | LCons of 'a * (unit -> 'a llist);;
You can pattern match like this:
let rec lfind e lxs =
match lxs with
| LNil -> false
| LCons(x, _) when x > e -> false
| LCons(x, xs) -> if e=x then true else lfind e (xs ())
;;
What would be the syntax (if possible at all) for returning the list of lists ([[a]]) but without the use of empty list ([]:[a])?
(similar as the second commented guard (2) below, which is incorrect)
This is a function that works correctly:
-- Split string on every (shouldSplit == true)
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith shouldSplit list = filter (not.null) -- would like to get rid of filter
(imp' shouldSplit list)
where
imp' _ [] = [[]]
imp' shouldSplit (x:xs)
| shouldSplit x = []:imp' shouldSplit xs -- (1) this line is adding empty lists
-- | shouldSplit x = [imp' shouldSplit xs] -- (2) if this would be correct, no filter needed
| otherwise = let (z:zs) = imp' shouldSplit xs in (x:z):zs
This is the correct result
Prelude> splitWith (== 'a') "miraaaakojajeja234"
["mir","koj","jej","234"]
However, it must use "filter" to clean up its result, so I would like to get rid of function "filter".
This is the result without the use of filter:
["mir","","","","koj","jej","234"]
If "| shouldSplit x = imp' shouldSplit xs" is used instead the first guard, the result is incorrect:
["mirkojjej234"]
The first guard (1) adds empty list so (I assume) compiler can treat the result as a list of lists ([[a]]).
(I'm not interested in another/different solutions of the function, just the syntax clarification.)
.
.
.
ANSWER:
Answer from Dave4420 led me to the answer, but it was a comment, not an answer so I can't accept it as answer. The solution of the problem was that I'm asking the wrong question. It is not the problem of syntax, but of my algorithm.
There are several answers with another/different solutions that solve the empty list problem, but they are not the answer to my question. However, they expanded my view of ways on how things can be done with basic Haskell syntax, and I thank them for it.
Edit:
splitWith :: (Char -> Bool) -> String -> [String]
splitWith p = go False
where
go _ [] = [[]]
go lastEmpty (x:xs)
| p x = if lastEmpty then go True xs else []:go True xs
| otherwise = let (z:zs) = go False xs in (x:z):zs
This one utilizes pattern matching to complete the task of not producing empty interleaving lists in a single traversal:
splitWith :: Eq a => (a -> Bool) -> [a] -> [[a]]
splitWith f list = case splitWith' f list of
[]:result -> result
result -> result
where
splitWith' _ [] = []
splitWith' f (a:[]) = if f a then [] else [[a]]
splitWith' f (a:b:tail) =
let next = splitWith' f (b : tail)
in if f a
then if a == b
then next
else [] : next
else case next of
[] -> [[a]]
nextHead:nextTail -> (a : nextHead) : nextTail
Running it:
main = do
print $ splitWith (== 'a') "miraaaakojajeja234"
print $ splitWith (== 'a') "mirrraaaakkkojjjajeja234"
print $ splitWith (== 'a') "aaabbbaaa"
Produces:
["mir","koj","jej","234"]
["mirrr","kkkojjj","jej","234"]
["bbb"]
The problem is quite naturally expressed as a fold over the list you're splitting. You need to keep track of two pieces of state - the result list, and the current word that is being built up to append to the result list.
I'd probably write a naive version something like this:
splitWith p xs = word:result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word:result,[])
else (result, x:word)
Note that this also leaves in the empty lists, because it appends the current word to the result whenever it detects a new element that satisfies the predicate p.
To fix that, just replace the list cons operator (:) with a new operator
(~:) :: [a] -> [[a]] -> [[a]]
that only conses one list to another if the original list is non-empty. The rest of the algorithm is unchanged.
splitWith p xs = word ~: result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word ~: result, [])
else (result, x:word)
x ~: xs = if null x then xs else x:xs
which does what you want.
I guess I had a similar idea to Chris, I think, even if not as elegant:
splitWith shouldSplit list = imp' list [] []
where
imp' [] accum result = result ++ if null accum then [] else [accum]
imp' (x:xs) accum result
| shouldSplit x =
imp' xs [] (result ++ if null accum
then []
else [accum])
| otherwise = imp' xs (accum ++ [x]) result
This is basically just an alternating application of dropWhile and break, isn't it:
splitWith p xs = g xs
where
g xs = let (a,b) = break p (dropWhile p xs)
in if null a then [] else a : g b
You say you aren't interested in other solutions than yours, but other readers might be. It sure is short and seems clear. As you learn, using basic Prelude functions becomes second nature. :)
As to your code, a little bit reworked in non-essential ways (using short suggestive function names, like p for "predicate" and g for a main worker function), it is
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = filter (not.null) (g list)
where
g [] = [[]]
g (x:xs)
| p x = [] : g xs
| otherwise = let (z:zs) = g xs
in (x:z):zs
Also, there's no need to pass the predicate as an argument to the worker (as was also mentioned in the comments). Now it is arguably a bit more readable.
Next, with a minimal change it becomes
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = case g list of ([]:r)-> r; x->x
where
g [] = [[]]
g (x:xs)
| p x = case z of []-> r; -- start a new word IF not already
_ -> []:r
| otherwise = (x:z):zs
where -- now z,zs are accessible
r#(z:zs) = g xs -- in both cases
which works as you wanted. The top-level case is removing at most one empty word here, which serves as a separator marker at some point during the inner function's work. Your filter (not.null) is essentially fused into the worker function g here, with the conditional opening1 of a new word (i.e. addition1 of an empty list).
Replacing your let with where allowed for the variables (z etc.) to became accessible in both branches of the second clause of the g definition.
In the end, your algorithm was close enough, and the code could be fixed after all.
1 when thinking "right-to-left". In reality the list is constructed left-to-right, in guarded recursion ⁄ tail recursion modulo cons fashion.
I have to define a function called zeros which takes input of two lists and returns a boolean which returns True if the number 0 appears the same amount of times in each list and false otherwise.
This is the last question in my homework and and I have managed to solve the question get it to work but I wondered if anybody can spot ways in which to reduce the amount of code, any ideas are appreciated. My code so far is as follows:
x :: Int
x = 0
instances::[Int]->Int
instances [] = 0
instances (y:ys)
| x==y = 1+(instances ys)
| otherwise = instances ys
zeros :: [Int] -> [Int] -> Bool
zeros [] [] = False
zeros x y
| ((instances x) == (instances y)) = True
| otherwise = False
Without giving too much away, since this is homework, here are a few hints.
Do you know about list comprehensions yet? They would be useful in this case. For example, you could combine them with an if expression to do something like this:
*Main> let starS s = [if c == 's' then '*' else ' ' | c <- s]
*Main> starS "schooners"
"* *"
You can even use them to do filtering. For example:
*Main> let findFives xs = [x | x <- xs, x == 5]
*Main> findFives [3,7,5,6,3,4,5,7,5,5]
[5,5,5,5]
Neither of these is a complete answer, but it shouldn't be hard to see how to adapt these structures to your situation.
You should also think about whether you actually need a guard here! For example, here's a function written with a guard in the same style as yours:
lensMatch [] [] = True
lensMatch xs ys
| ((length xs) == (length ys)) = True
| otherwise = False
Here's a function that does the same thing!
lensMatch' xs ys = length xs == length ys
You can see that they are the same; testing the first:
*Main> lensMatch [1..4] [1..4]
True
*Main> lensMatch [1..4] [1..5]
False
*Main> lensMatch [] [1..5]
False
*Main> lensMatch [] []
True
And testing the second:
*Main> lensMatch' [1..4] [1..4]
True
*Main> lensMatch' [1..4] [1..5]
False
*Main> lensMatch' [] [1..5]
False
*Main> lensMatch' [] []
True
Finally, I agree very strongly with sblom's comment above; zeros [] [] should be True! Think about the following statement: "For each item x in set s, x > 0". If set s is empty, then the statement is true! It's true because there are no items in s at all. This seems to me like a similar situation.
I can't believe nobody has suggested to use foldr yet. Not the shortest or best definition, but IMO the most educational:
instances :: Eq a => a -> [a] -> Int
instances n = foldr incrementIfEqual 0
where incrementIfEqual x subtotal
| x == n = subtotal + 1
| otherwise = subtotal
zeros :: Num a => [a] -> [a] -> Bool
zeros xs ys = instances 0 xs == instances 0 ys
Though for a really brief definition of instances, what I came up with is basically the same as Abizern:
instances :: Eq a => a -> [a] -> Int
instances x = length . filter (==x)
Have you thought of doing this in one pass by filtering each list to get just the zeroes and then comparing the length of the lists to see if they are equal?
zeroCompare xs ys = cZeroes xs == cZeroes ys
where
cZeroes as = length $ filter (== 0) as
Instead of length and filter, you can take the result of a predicate p, convert it to 0 or 1, and sum the result:
count p = sum . map (fromEnum.p)
--or
import Data.List
count p = foldl' (\x -> (x+).fromEnum.p) 0
In your case, p is of course (==0). Converting Bool to Int using fromEnum is a very useful trick.
Another idea would be to deal with both list simultaneously, which is a little bit lengthy, but easy to understand:
zeros xs ys = cmp xs ys == 0 where
cmp (0:xs) ys = cmp xs ys + 1
cmp xs (0:ys) = cmp xs ys - 1
cmp (_:xs) ys = cmp xs ys
cmp xs (_:ys) = cmp xs ys
cmp [] [] = 0
I would break the problem down into smaller problems involving helper functions.
This is how I would break it down:
Main function to compare two counts
Count helper function
First: You need a way to count the amount of zeroes in a list. For example, I would approach this by doing the following if searching for the number of 0 in an integer list:
count :: [Int] -> Int
count xs = foldl (\count num -> if num == 0 then (count + 1) else count) 0 xs
Second: You need a way to compare the count of two lists. Essentially, you need a function that takes two lists in as parameters, calculates the count of each list, and then returns a boolean depending on the result. For example, if each list is an int list, corresponding with my count example above:
equalZeroes :: [Int] -> [Int] -> Bool
equalZeroes x y = (count x) == (count y)
You could also define count under the where keyword inside the equalZeroes function like so:
equalZeroes :: [Int] -> [Int] -> Bool
equalZeroes x y = (count x) == (count y)
where
count :: [Int] -> Int
count xs = foldl (\count num -> if num == 0 then (count + 1) else count) 0 xs
When running this code, calling the function as so would get the desired boolean values returned:
equalZeroes [0,1,4,5,6] [1,4,5,0,0]
-> False
equalZeroes [0,1,4,5,6] [1,4,5,0]
-> True